**Q: Let f(x) = 3x² – x**

**a. Find the average rate of change of f(x) with the respect to x as x changes from x = 0 to x = 1/16**

**b. Use calculus to find the instantaneous rate of change of f(x) at x=0 and compare with the average rate found in part (a).**

A: Let’s start with part (a). The **average rate of change **is nothing more than a slope of a line. Remember,

Slope = m = (y_{2} – y_{1}) / (x_{2} – x_{1})

Well, we know x_{1} = 0 and x_{2} = 1/16… How do we find y_{1} and y_{2}? We plug each x into the original function to find the corresponding y value, like so:

f(x) = 3x² – x

f(0) = 3(0)² – 0 = 0

So, the point** (x _{1} , y_{1}) = (0, 0)**

Now try for x = 1/16:

f(1/16) = 3(1/16)² – 1/16 = 3(1/256) – 1/16

= 3/256 – 1/16 = -13/256

So, the point **(x _{2} , y_{2}) = (1/16, -13/256)**

Now, calculate the **average rate of change (slope):**

m = (y_{2} – y_{1}) / (x_{2} – x_{1}) = (-13/256 – 0) / (1/16 – 0) = (-13/256) / (1/16)

[remember, diving by a fraction is the same as multiplying by its reciprocal!]

= -13/256 * 16/1 = -13/16

The average rate of change from x = 0 to x = 1/16 is **-13/16**

(b) Now, the **instantaneous rate of change** is asking for the rate of change at a single point. In our case, the specific point is when x = 0. We need calculus to find the rate of change at a single point….. Remember, the **derivative **gives us the rate of change (slope) at a single point.

So, take the derivative first:

f(x) = 3x² – x

f ‘ (x) = 6x – 1

Now, we are curious about the point where x = 0… So, plug in 0 for x:

f ‘ (x) = 6(0) – 1 = -1

The **instantaneous rate of change **of the function when x = 0 is **-1.**

How does this answer compare the the answer in part (a)? Well, an average rate of change of -13/16 is pretty darn close to the instantaneous rate of change which was -1. The average rate of change gives a pretty good estimate of the instantaneous rate of change, as long as you are on a continuous function and use two points that are pretty close together.

A very good explanation for someone like myself who has just started this topic at school.

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